# ANGLE SUBTENDED BY AN ARC AT THE CENTER OF A CIRCLE

Theorem :

When two angles subtended by the same arc, the angle at the center of the circle is twice the angle at the circumference

Considering the circle with center O, now placing the points A, B and C on the circumference.

∠ACO = ∠OAC

∠BCO = ∠OBC

∠AOD = Exterior angle of the triangle

∠AOD = ∠ACO + ∠OAC  ----(1)

∠BOD = ∠BCO + ∠OBC  ----(2)

(1) + (2)

∠AOD + ∠BOD = (∠ACO + ∠OAC) + (∠BCO + ∠OBC)

∠AOB = (∠ACO + ∠OAC) + (∠BCO + ∠OBC)

∠AOB = 2∠OCA + 2∠OCB

∠AOB = 2(∠OCA + ∠OCB)

∠AOB = 2∠ACB

Work out the size of each angle marked with a letter. Give reasons for your answers.

Problem 1 :

Solution :

AOC + 230˚ = 360˚

AOC = 360 - 230

AOC = 130˚

a = 130˚

b = 65˚

Since ABCD is a cyclic quadrilateral, the opposite angles add upto 180.

ABC + 65 = 180˚

ABC = 180 - 65

ABC = 115˚

c = 115˚

Problem 2 :

Solution :

AOC = 360 - 106 = 254˚

2ABC = reflex of AOC

Reflex of AOC = 2g

g = reflex of AOC /2

= 254/2

= 127˚

So, the value of g is 127˚.

Problem 3 :

P, Q and R are points on the circumference of a circle, center, O. Angle PRQ = 64˚. SP and SQ are tangents to the circle at the points P and Q respectively.

Work out the size of angle

i) PSQ      ii) PQO       iii) POS      iv) QSO

Solution :

POQ = 2 PRQ

POQ = 2(64˚)

POQ = 128˚

i) PSQ :

PSQ + POQ = 180˚

PSQ + 128 = 180

PSQ = 180 - 126

PSQ = 52˚

∠OSP = 52˚/2 ==> 26

In triangle PSQ,

SPQ + ∠SQP + PSQ = 180

Here SPQ = ∠SQP (SP and SQ are tangents drawn from the external point, they will have same length).

2SPQ + 52 = 180

SPQ = 64

ii) PQO :

PQO = ∠SQO - ∠SQP

PQO = 90 - 64

PQO = 26

iii) POS :

∠OPS + ∠OSP + ∠POS = 180

90 + 26 ∠POS = 180

∠POS = 180 - 116

∠POS = 64

iv) QSO :

QSO = 1/2 PSQ

= 1/2(52)

QSO = 26˚

Problem 4 :

P, Q and R are points on the circumference of a circle, center, O. Angle PSQ = 60˚. SP and SQ are tangents to the circle at the points P and Q respectively.

a. Work out the size of angle

i) QPS      ii) PQO      iii) PRQ      iv) POQ

b. what type of triangle is PQS?

c. Given that angle OQR = 10, work out the size of angle OPR.

Solution :

a. Given, SP = SQ

i) QPS :

PSQ = 60˚, so QPS = PQS

QPS = PQS = 1/2(180 - 60)

QPS = 60˚

ii) PQO :

PQO = SQO - SQP

= 90˚ - 60˚

PQO = 30˚

iii) POQ :

POQ = 360˚ - (PSQ + SPO + SQO)

= 360˚ - (60 + 90 + 90)

= 360 - 240

POQ = 120˚

iv) PRQ :

PRQ = 1/2 POQ

= 1/2(120)

PRQ = 60˚

b. Given, PSQ = 60˚

SP = SQ

So, PSQ is an equilateral triangle.

c. OQR = 10˚

RQP = OQR + OQP

= 10˚ + 30˚ = 40˚

RPQ = 180˚ - RQP - QRP

= 180 - 40 - 60 = 80˚

QPO = SPO - SPQ

= 180 - 60 = 30˚

OPR = RPQ - SPO

OPR = 80 - 30

OPR = 50˚

Problem 5 :

P, Q, R and S are points on the circumference of a circle, center, O. PT and TR are tangents to the circle. OST is a straight line.

Angle OTR = 38˚

Find the size of the angle:

i) ROT  ii) PQR  iii) SRT  iv) PSO  v) PST

Solution :

i) ROT :

TPO = TRO = 90˚

90 + 38 + ROT = 180˚

128 + ROT = 180

ROT = 180 - 128

ROT = 52˚

ii) PQR :

POR = 2 ROT

= 2(52˚)

POR = 104˚

PQR = 1/2 POR

= 1/2 (104˚)

PQR = 52˚

iii) SRT :

OR = OS

ORS = OSR

RSO = 1/2(ORS + RSO)

= 1/2(180 - 52)

RSO = 64˚

SRT = RSO - OTR

SRT = 64˚ - 38˚

SRT = 26˚

iv) PSO :

PSO = RSO = 64˚

v) PST :

PST = RST

= 180˚ - RSO

= 180˚ - 64˚

PST = 116˚

## Recent Articles

1. ### Finding Range of Values Inequality Problems

May 21, 24 08:51 PM

Finding Range of Values Inequality Problems

2. ### Solving Two Step Inequality Word Problems

May 21, 24 08:51 AM

Solving Two Step Inequality Word Problems